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MATH 4171 Graph Theory
SPRING 2006
Homework Set I  Solutions
1.
Describe a real world problem (preferable from your own ﬁeld) that can be modeled by graphs.
Specify vertices, edges, and the incidence relation for your graph.
...
2.
Is there a
simple
graph on 2006 vertices for which no two vertices have the same degree? You
need to justify your conclusion.
Answer:
No.
Justiﬁcation.
Suppose there was such a graph
G
= (
V, E
). Then for each
v
∈
V
, since
G
is simple,
v
does not join to itself and there is at most one edge between
v
and any other vertex. It follows
that 0
≤
d
G
(
v
)
≤
2005, for all
v
∈
V
. Since
G
has 2006 vertices and there are only 2006 choices
for the degrees, the choice of
G
implies that very number in
{
0
,
1
,
2
, ...,
2005
}
would be the degree
of some vertex. In particular,
G
has a vertex
x
with
d
G
(
x
) = 0 and a vertex
y
with
d
G
(
y
) = 2005.
Clearly, this is impossible since either
xy
∈
E
, which would imply
d
G
(
x
)
≥
1 and would contradict
d
G
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 Spring '08
 Lax,R
 Graph Theory

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